AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that (i) AD bisects BC (ii) AD bisects A.

Solution

Given

 AD is an altitude and AB = AC

To Prove

(i) AD bisects BC (ii) AD bisects A.

Proof

∆ADB and ∆ADC

∠ADB=∠ADC ——– ———–[each 90°]

AB=AC ——————–[given]

AD=AD ——–[common]

∴ ∆ADB ≅∆ADC

BD=DC ————-[c.p.c.t]

∴AD bisects BC

∠1=∠2 ————-[c.p.c.t]

∴AD bisects ∠A

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